If $k$ has a factor of square number greater than $1$, let $t^2 \mid k$, $t > 1$, then taking $m = n = k - \frac{k}{t} + 1$, we see that such $k$ has the properties.

If $k$ has no factor of square number, if there are two primes $p_1, p_2$ such that $(p_1-2)(p_2-2) \ge 4$ and $p_1 p_2 \mid k$. Let $k = p_1 p_2 \cdots p_r$ and $p_1, p_2, \dots, p_r$ be pairwise different, $r \ge 2$. Since there is at least one of $(p_1-1)p_2 p_3 \cdots p_r + 1$ and $(p_1-2)p_2 p_3 \cdots p_r + 1$ is coprime with $p_1$ (otherwise $p_1$ divides their difference $p_2 p_3 \cdots p_r$, which is a contradiction), taking this number as the number $m$, then, $1 < m < k$, $(m, k) = 1$. Similarly, we can take a number $n$ of $(p_2-1)p_1 p_3 \cdots p_r + 1$ or $(p_2-2)p_1 p_3 \cdots p_r + 1$ such that $1 < n < k$, $(n, k) = 1$. So $p_1 p_2 \cdots p_r \mid (m-1)(n-1)$, and
$$m + n \ge (p_1 - 2)p_2 p_3 \cdots p_r + 1 + (p_2 - 2)p_1 p_3 \cdots p_r + 1 \\ = k + ((p_1 - 2)(p_2 - 2) - 4)p_3 \cdots p_r + 2 > k.$$
Such $m, n$ will satisfy the conditions.

If there are no two primes $p_1, p_2$ such that $(p_1-2)(p_2-2) \ge 4$, $p_1 p_2 \mid k$, then it is easy to verify such integer $k \ge 3$ can only be $15$, $30$ or as $p$, $2p$ (where $p$ is an odd prime). It is easy to see that, if $k = p, 2p, 30$, then there are no $m, n$ satisfying the conditions; if $k = 15$, then $m = 11, n = 13$ satisfy the conditions.

Summing up, integer $k \ge 3$ satisfies the conditions if and only if $k$ is not an odd prime, nor double of an odd prime and nor $30$.